Question 188598
y^2-8x-4y+12=0
Isolate the 'x' to one side of the equation:
y^2-4y+12 = 8x
(1/8)y^2-(1/2)y+(3/2) = x
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Because 'y' is squared it will be a HORIZONTAL parabola.
Because the coefficient associated with the y^2 term is positive: open right
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Completing the square:
(1/8)(y^2-4y)+(3/2) = x
(1/8)(y^2-4y+4)+(3/2 - 1/2) = x
(1/8)(y^2-4y+4) + 1 = x
(1/8)(y-2)^2 + 1 = x
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This now is in the form of:
x = (1/(4c))(y-k)^2 + h
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From the above, we see that
vertex = (1,2)
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c = distance between vertex and the focus/directrix
1/(4c) = 1/8
cross-multiplying:
4c = 8
c = 2
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focus is at (h+c,k)
focus = (1+2, 2) = (3, 2)
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directrix is at:
x = h-c 
x = 1-2
x = -1